Question

A string is under tension so that its length is increased by $\frac{1}{n}$ times its original length. The ratio of fundamental frequency of longitudinal vibrations and transverse vibrations will be

• A. $1:1/\sqrt{(}1+1/n)$
• B. $n^2:1$
• C. $\sqrt{n}:1$
• D. $n:1$

Answer 1

The correct option is A $1:1/\sqrt{(}1+1/n)$

Let the initial length be $=l_0$

Velocity of the wave $V_0=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{{Tl}_0}{m}}$

Frequency of $f_0=\frac{V_0}{\lambda }=\frac{V_0}{2l_0}$

New length $=l_0(1+\frac{1}{n})$

$\lambda =2l_0(1+\frac{1}{n})$

New velocity $V=\sqrt{\frac{T\times l_0(1+1/n)}{m}}=\sqrt{(1+\frac{1}{n})V_0}$

Frequency $f=\frac{\sqrt[V_0]{V_0(1+\frac{1}{n})}}{2l_0(1+\frac{1}{n})}$

$[F_T=\sqrt{(1+\frac{1}{n})}\sqrt{\frac{T{l}_0}{m}}]$ fundamental freq of transverse wave vibration

for longitudinal wave

initial velocity $V_0=\sqrt{\frac{Y}{S}}=\sqrt{\frac{{Yl}_0}{K}}$

consider density $S=m(l\times A)$

Hence density is inversely proportional to length

$S=\left(\frac{k}{l}\right)$ where $l$ is constant.

New velocity $V=\sqrt{\frac{Y\times l_0(1+1/n)}{K}}={\left(\sqrt{1+\frac{1}{n}}\right)}^{V_6}$

$f_0=\frac{V_0}{{2l}_0}=\frac{1}{2l_0}\sqrt{\frac{Y{l}_0}{K}}$

frequency of wave $f=\frac{V_0\left(\sqrt{1+1/n}\right)}{2l_0(1+1/n)}$

Hence,

$[\frac{n}{n_0}=\frac{1}{\sqrt{1+1/n}}]$

$\therefore$ Option (A) is correct.